\(\int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx\) [249]

Optimal result

Integrand size = 25, antiderivative size = 197 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {3 \text {arctanh}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}} \] Output:

exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)-2*exp(c*(b*x+ 
a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^2/(tanh(b*c*x+a*c)^2)^(1/2)+3 
*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))/(tanh(b*c*x+a*c)^ 
2)^(1/2)-3*arctanh(exp(c*(b*x+a)))*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2) 
^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 7.14 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.70 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=-\frac {e^{-5 c (a+b x)} \left (-21 \left (252105+507305 e^{2 c (a+b x)}+173916 e^{4 c (a+b x)}-154296 e^{6 c (a+b x)}-73885 e^{8 c (a+b x)}+4887 e^{10 c (a+b x)}\right )-\frac {315 \left (-16807-28218 e^{2 c (a+b x)}+1173 e^{4 c (a+b x)}+17748 e^{6 c (a+b x)}+4299 e^{8 c (a+b x)}-1434 e^{10 c (a+b x)}+7 e^{12 c (a+b x)}\right ) \text {arctanh}\left (\sqrt {e^{2 c (a+b x)}}\right )}{\sqrt {e^{2 c (a+b x)}}}+384 e^{8 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^2 \left (7+5 e^{2 c (a+b x)}\right ) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )+256 e^{8 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )\right ) \tanh ^3(c (a+b x))}{60480 b c \tanh ^2(c (a+b x))^{3/2}} \] Input:

Integrate[E^(c*(a + b*x))/(Tanh[a*c + b*c*x]^2)^(3/2),x]
 

Output:

-1/60480*((-21*(252105 + 507305*E^(2*c*(a + b*x)) + 173916*E^(4*c*(a + b*x 
)) - 154296*E^(6*c*(a + b*x)) - 73885*E^(8*c*(a + b*x)) + 4887*E^(10*c*(a 
+ b*x))) - (315*(-16807 - 28218*E^(2*c*(a + b*x)) + 1173*E^(4*c*(a + b*x)) 
 + 17748*E^(6*c*(a + b*x)) + 4299*E^(8*c*(a + b*x)) - 1434*E^(10*c*(a + b* 
x)) + 7*E^(12*c*(a + b*x)))*ArcTanh[Sqrt[E^(2*c*(a + b*x))]])/Sqrt[E^(2*c* 
(a + b*x))] + 384*E^(8*c*(a + b*x))*(1 + E^(2*c*(a + b*x)))^2*(7 + 5*E^(2* 
c*(a + b*x)))*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*c 
*(a + b*x))] + 256*E^(8*c*(a + b*x))*(1 + E^(2*c*(a + b*x)))^3*Hypergeomet 
ricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, E^(2*c*(a + b*x))])*Tanh[ 
c*(a + b*x)]^3)/(b*c*E^(5*c*(a + b*x))*(Tanh[c*(a + b*x)]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.54, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7271, 2720, 25, 300, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^3(a c+b x c)dx}{\sqrt {\tanh ^2(a c+b c x)}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\tanh (a c+b c x) \int -\frac {\left (1+e^{2 c (a+b x)}\right )^3}{\left (1-e^{2 c (a+b x)}\right )^3}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\tanh (a c+b c x) \int \frac {\left (1+e^{2 c (a+b x)}\right )^3}{\left (1-e^{2 c (a+b x)}\right )^3}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\)

\(\Big \downarrow \) 300

\(\displaystyle -\frac {\tanh (a c+b c x) \int \left (\frac {2 \left (1+3 e^{4 c (a+b x)}\right )}{\left (1-e^{2 c (a+b x)}\right )^3}-1\right )de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (-3 \text {arctanh}\left (e^{c (a+b x)}\right )+e^{c (a+b x)}+\frac {3 e^{c (a+b x)}}{1-e^{2 c (a+b x)}}-\frac {2 e^{c (a+b x)}}{\left (1-e^{2 c (a+b x)}\right )^2}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}\)

Input:

Int[E^(c*(a + b*x))/(Tanh[a*c + b*c*x]^2)^(3/2),x]
 

Output:

((E^(c*(a + b*x)) - (2*E^(c*(a + b*x)))/(1 - E^(2*c*(a + b*x)))^2 + (3*E^( 
c*(a + b*x)))/(1 - E^(2*c*(a + b*x))) - 3*ArcTanh[E^(c*(a + b*x))])*Tanh[a 
*c + b*c*x])/(b*c*Sqrt[Tanh[a*c + b*c*x]^2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.68 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {\operatorname {csgn}\left (\tanh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\cosh \left (b c x +a c \right )^{2}}{\sinh \left (b c x +a c \right )}-\frac {2}{\sinh \left (b c x +a c \right )}+\frac {\cosh \left (b c x +a c \right )^{3}}{\sinh \left (b c x +a c \right )^{2}}-\frac {3 \cosh \left (b c x +a c \right )}{\sinh \left (b c x +a c \right )^{2}}+\frac {3 \,\operatorname {csch}\left (b c x +a c \right ) \coth \left (b c x +a c \right )}{2}-3 \,\operatorname {arctanh}\left ({\mathrm e}^{b c x +a c}\right )\right )}{c b}\)	\(131\)
risch	\(\frac {2 \,{\mathrm e}^{5 c \left (b x +a \right )}+3 \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right ) {\mathrm e}^{4 c \left (b x +a \right )}-3 \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right ) {\mathrm e}^{4 c \left (b x +a \right )}-10 \,{\mathrm e}^{3 c \left (b x +a \right )}-6 \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right ) {\mathrm e}^{2 c \left (b x +a \right )}+6 \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right ) {\mathrm e}^{2 c \left (b x +a \right )}+4 \,{\mathrm e}^{c \left (b x +a \right )}+3 \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )-3 \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, c b}\)	\(211\)

Input:

int(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

csgn(tanh(c*(b*x+a)))/c/b*(1/sinh(b*c*x+a*c)*cosh(b*c*x+a*c)^2-2/sinh(b*c* 
x+a*c)+cosh(b*c*x+a*c)^3/sinh(b*c*x+a*c)^2-3/sinh(b*c*x+a*c)^2*cosh(b*c*x+ 
a*c)+3/2*csch(b*c*x+a*c)*coth(b*c*x+a*c)-3*arctanh(exp(b*c*x+a*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (179) = 358\).

Time = 0.11 (sec) , antiderivative size = 613, normalized size of antiderivative = 3.11 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")
 

Output:

1/2*(2*cosh(b*c*x + a*c)^5 + 10*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 + 2* 
sinh(b*c*x + a*c)^5 + 10*(2*cosh(b*c*x + a*c)^2 - 1)*sinh(b*c*x + a*c)^3 - 
 10*cosh(b*c*x + a*c)^3 + 10*(2*cosh(b*c*x + a*c)^3 - 3*cosh(b*c*x + a*c)) 
*sinh(b*c*x + a*c)^2 - 3*(cosh(b*c*x + a*c)^4 + 4*cosh(b*c*x + a*c)*sinh(b 
*c*x + a*c)^3 + sinh(b*c*x + a*c)^4 + 2*(3*cosh(b*c*x + a*c)^2 - 1)*sinh(b 
*c*x + a*c)^2 - 2*cosh(b*c*x + a*c)^2 + 4*(cosh(b*c*x + a*c)^3 - cosh(b*c* 
x + a*c))*sinh(b*c*x + a*c) + 1)*log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) 
 + 1) + 3*(cosh(b*c*x + a*c)^4 + 4*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^3 + 
 sinh(b*c*x + a*c)^4 + 2*(3*cosh(b*c*x + a*c)^2 - 1)*sinh(b*c*x + a*c)^2 - 
 2*cosh(b*c*x + a*c)^2 + 4*(cosh(b*c*x + a*c)^3 - cosh(b*c*x + a*c))*sinh( 
b*c*x + a*c) + 1)*log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) - 1) + 2*(5*co 
sh(b*c*x + a*c)^4 - 15*cosh(b*c*x + a*c)^2 + 2)*sinh(b*c*x + a*c) + 4*cosh 
(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^4 + 4*b*c*cosh(b*c*x + a*c)*sinh(b*c 
*x + a*c)^3 + b*c*sinh(b*c*x + a*c)^4 - 2*b*c*cosh(b*c*x + a*c)^2 + 2*(3*b 
*c*cosh(b*c*x + a*c)^2 - b*c)*sinh(b*c*x + a*c)^2 + b*c + 4*(b*c*cosh(b*c* 
x + a*c)^3 - b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c))
 

Sympy [F]

\[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=e^{a c} \int \frac {e^{b c x}}{\left (\tanh ^{2}{\left (a c + b c x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)**2)**(3/2),x)
 

Output:

exp(a*c)*Integral(exp(b*c*x)/(tanh(a*c + b*c*x)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.57 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=-\frac {3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{2 \, b c} + \frac {3 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} - 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} - 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \] Input:

integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")
 

Output:

-3/2*log(e^(b*c*x + a*c) + 1)/(b*c) + 3/2*log(e^(b*c*x + a*c) - 1)/(b*c) + 
 (e^(5*b*c*x + 5*a*c) - 5*e^(3*b*c*x + 3*a*c) + 2*e^(b*c*x + a*c))/(b*c*(e 
^(4*b*c*x + 4*a*c) - 2*e^(2*b*c*x + 2*a*c) + 1))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.89 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=\frac {e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} - \frac {3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{2 \, b c} + \frac {3 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{2 \, b c} - \frac {3 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2}} \] Input:

integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")
 

Output:

e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) - 3/2*log(e^(b*c*x + a* 
c) + 1)*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) + 3/2*log(abs(e^(b*c*x + a*c) - 
 1))*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) - (3*e^(3*b*c*x + 3*a*c)*sgn(e^(2* 
b*c*x + 2*a*c) - 1) - e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(b*c*( 
e^(2*b*c*x + 2*a*c) - 1)^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2}} \,d x \] Input:

int(exp(c*(a + b*x))/(tanh(a*c + b*c*x)^2)^(3/2),x)
 

Output:

int(exp(c*(a + b*x))/(tanh(a*c + b*c*x)^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.54 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx=\frac {e^{3 b c x +3 a c} \tanh \left (b c x +a c \right )^{2}+2 e^{3 b c x +3 a c} \tanh \left (b c x +a c \right )-e^{3 b c x +3 a c}-3 e^{2 b c x +2 a c} \mathrm {log}\left (e^{b c x +2 a c}+e^{a c}\right ) \tanh \left (b c x +a c \right )^{2}+3 e^{2 b c x +2 a c} \mathrm {log}\left (e^{b c x +2 a c}-e^{a c}\right ) \tanh \left (b c x +a c \right )^{2}-7 e^{b c x +a c} \tanh \left (b c x +a c \right )^{2}-2 e^{b c x +a c} \tanh \left (b c x +a c \right )+e^{b c x +a c}+3 \,\mathrm {log}\left (e^{b c x +2 a c}+e^{a c}\right ) \tanh \left (b c x +a c \right )^{2}-3 \,\mathrm {log}\left (e^{b c x +2 a c}-e^{a c}\right ) \tanh \left (b c x +a c \right )^{2}}{2 \tanh \left (b c x +a c \right )^{2} b c \left (e^{2 b c x +2 a c}-1\right )} \] Input:

int(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(3/2),x)
 

Output:

(e**(3*a*c + 3*b*c*x)*tanh(a*c + b*c*x)**2 + 2*e**(3*a*c + 3*b*c*x)*tanh(a 
*c + b*c*x) - e**(3*a*c + 3*b*c*x) - 3*e**(2*a*c + 2*b*c*x)*log(e**(2*a*c 
+ b*c*x) + e**(a*c))*tanh(a*c + b*c*x)**2 + 3*e**(2*a*c + 2*b*c*x)*log(e** 
(2*a*c + b*c*x) - e**(a*c))*tanh(a*c + b*c*x)**2 - 7*e**(a*c + b*c*x)*tanh 
(a*c + b*c*x)**2 - 2*e**(a*c + b*c*x)*tanh(a*c + b*c*x) + e**(a*c + b*c*x) 
 + 3*log(e**(2*a*c + b*c*x) + e**(a*c))*tanh(a*c + b*c*x)**2 - 3*log(e**(2 
*a*c + b*c*x) - e**(a*c))*tanh(a*c + b*c*x)**2)/(2*tanh(a*c + b*c*x)**2*b* 
c*(e**(2*a*c + 2*b*c*x) - 1))